Apply qualitatively and quantitatively Newton's Law of Universal Gravitation, F = G m_1 m_2 / r^2, to determine the magnitude of force, gravitational field strength g = G M / r^2, and acceleration due to gravity at different points in a radial gravitational field

What this dot point is asking

NESA wants you to apply Newton's Law of Universal Gravitation to calculate the force between two masses, determine the gravitational field strength at a point, and explain how the inverse-square dependence on distance shapes planetary and satellite motion. You need both the conceptual explanation (action at a distance, field model) and the numerical fluency to compute $F$ and $g$ at arbitrary distances.

The answer

Newton's Law of Universal Gravitation

Every pair of point masses attracts each other with a force directed along the line joining them:

where:

• $G = 6.67 \times 10^{-11}$ N m$^2$/kg$^2$ is the universal gravitational constant.
• $m_1$ and $m_2$ are the two masses in kilograms.
• $r$ is the distance between their centres (not surfaces).

The force is mutual: each mass exerts the same magnitude of force on the other (Newton's third law).

The inverse-square law

Force is inversely proportional to the square of the distance. Doubling $r$ reduces the force to one quarter. Halving $r$ quadruples the force. This rapid fall-off explains why Earth's gravity dominates near the surface but becomes negligible far from the planet.

Gravitational field strength

The gravitational field strength $g$ at a point is the gravitational force per unit mass on a test mass placed there:

where $M$ is the mass of the source body and $r$ is the distance from its centre. Units: N/kg or m/s$^2$ (numerically equal).

At Earth's surface ($r = R_E = 6.37 \times 10^6$ m): $g \approx 9.8 \text{ m/s}^2$.

Acceleration due to gravity

For an object of mass $m$ in free fall in a gravitational field $g$, the acceleration is $a = g$ (regardless of $m$, because $F = mg$ and $a = F/m$). All objects fall with the same acceleration in a given gravitational field, in the absence of air resistance.

Field model versus action at a distance

Two equivalent descriptions:

• Action at a distance: the two masses pull on each other directly across empty space.
• Field model: each mass creates a gravitational field around itself, and any other mass in that field experiences a force. The field model is preferred for HSC because it generalises cleanly to electric and magnetic fields.

Examples in context

Example 1. Gravitational pull on the Parkes 64 m dish. The Parkes radio telescope dish has a mass of $300{,}000 \text{ kg}$. At Earth's surface ($M_E = 5.97 \times 10^{24} \text{ kg}$, $r_E = 6.37 \times 10^6 \text{ m}$), Newton's law gives $F = G M_E m / r_E^2 = 6.674 \times 10^{-11} \times 5.97 \times 10^{24} \times 3.0 \times 10^5 / (6.37 \times 10^6)^2 = 2.94 \times 10^6 \text{ N}$. This is exactly $m g = 3 \times 10^5 \times 9.80 = 2.94 \times 10^6 \text{ N}$ as expected, confirming that the surface field strength $g = G M_E / r_E^2 = 9.80 \text{ m/s}^2$ is the same for every mass.

Example 2. Moon-Earth attraction and ocean tides on the NSW coast. With Earth-Moon distance $r = 3.84 \times 10^8 \text{ m}$ and $M_{\text{Moon}} = 7.35 \times 10^{22} \text{ kg}$, the Moon's gravitational acceleration at Earth's centre is $g_M = G M_{\text{Moon}} / r^2 = 3.32 \times 10^{-5} \text{ m/s}^2$. The tidal effect along the Coffs Harbour tide gauge comes from the difference in $g_M$ between Earth's near side and centre: $\Delta g = 2 G M_{\text{Moon}} r_E / r^3 = 1.10 \times 10^{-6} \text{ m/s}^2$. This $\sim 10^{-7} g$ differential pulls the ocean into the tidal bulges that NSW Ports records as roughly $1.5 \text{ m}$ semidiurnal tides.

Try this

Q1. State Newton's law of universal gravitation in words and as an equation, defining every symbol. [2 marks]

• Cue. Word statement + $F = G m_1 m_2 / r^2$ with $G = 6.674 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$, $r$ between centres.

Q2. Calculate the gravitational field strength on the surface of Mars given $M = 6.42 \times 10^{23} \text{ kg}$ and $r = 3.39 \times 10^6 \text{ m}$. [2 marks]

• Cue. $g = G M / r^2 = 6.674 \times 10^{-11} \times 6.42 \times 10^{23} / (3.39 \times 10^6)^2 = 3.73 \text{ m/s}^2$.

Q3. A satellite of mass $500 \text{ kg}$ is placed at an altitude of $400 \text{ km}$ above Earth. (a) Calculate the gravitational force on it. (b) Compare this with its weight at Earth's surface. (c) Explain why this is the centripetal force keeping the satellite in orbit. [2+2+2 marks]

• Cue. (a) $r = 6.77 \times 10^6 \text{ m}$, $F = G M_E m / r^2 = 4340 \text{ N}$. (b) Surface $W = 4900 \text{ N}$, so $\sim 89\%$. (c) Only force acting; directed toward centre, supplies $m v^2 / r$.